HW5 - Solutions to Homework 5 Math 293 Section 2.3 Problem...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Homework # 5, Math 293 Section: 2.3 Problem 19 Acceleration due to the motor= 4 ft/s 2 Deceleration due to water resistance = v 2 / 400 ft/s 2 So, the differential equation is: v = 4- 1 400 v 2 , v (0) = 0 or Z dv 4- 1 400 v 2 = Z dt Z 1 40 1- ( v/ 40) 2 = Z 1 10 dt tanh- 1 ( v 40 ) = t/ 10 + C at t = 0, v = 0. Therefore, C = 0. v ( t ) = 40 tanh( t 10 ) v (10) = 40 tanh(1) = 30 . 46 ft/s v ( ∞ ) = 40 tanh( ∞ ) = 40 . Problem 28 Given differential equation is dv dt =- GM r 2 = v dv dr =- GM r 2 After integration and using the initial conditions r (0) = r , v (0) = 0, we have v 2 2 = GM 1 r- 1 r 1 dr dt = v =- s 2 GM 1 r- 1 r (1) We take the negative square root because v < 0 in descent. Hence, t =- r r 2 GM Z s r r- r dr using r = r cos 2 θ , t = r r 2 GM Z 2 r cos 2 θdθ t = r 3 / 2 √ 2 GM Z (1 + cos 2 θdθ ) t = r 3 / 2 √ 2 GM ( θ + 1 2 sin 2 θ ) + C or t = r 3 / 2 √ 2 GM ( θ + sin θ cos θ ) + C Using the initial condition t = 0, r = r and θ (0) = 0, we get...
View Full Document

This note was uploaded on 04/01/2008 for the course MATH 2930 taught by Professor Terrell,r during the Fall '07 term at Cornell.

Page1 / 4

HW5 - Solutions to Homework 5 Math 293 Section 2.3 Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online